Question: You have found the following ages (in years) of 5 zebras. Those zebras were randomly selected from the 38 zebras at your local zoo: $ 25,\enspace 10,\enspace 4,\enspace 21,\enspace 1$ Based on your sample, what is the average age of the zebras? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we only have data for a small sample of the 38 zebras, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $5$ samples and divide by $5$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\overline{x}} = \dfrac{25 + 10 + 4 + 21 + 1}{{5}} = {12.2\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {163.84} + {4.84} + {67.24} + {77.44} + {125.44}} {{5 - 1}} $ {s^2} = \dfrac{{438.8}}{{4}} = {109.7\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{109.7\text{ years}^2}} = {10.5\text{ years}} $ We can estimate that the average zebra at the zoo is 12.2 years old. There is also a standard deviation of 10.5 years.